A point positive charge of Q' units is moved round another point positive charge of Q units in circular path. If the radius of the circle r is the work done on the charge Q' in making one complete revolution i-

Options

A. `(Q)/(4pi in_(0) r)`
B. `(Q Q')/(4pi in_(0) r)`
C. `(Q')/(4pi in_(0) r)`
D. 0

Answer

To find the work done on a point positive charge \( Q' \) when it is moved around another point positive charge \( Q \) in a circular path of radius \( r \), we can follow these steps:

### Step 1: Understand the Forces Involved
When a positive charge \( Q' \) is placed near another positive charge \( Q \), they repel each other due to the electrostatic force. The force \( F \) between the two charges can be calculated using Coulomb's law:
\[ F = k \frac{Q \cdot Q'}{r^2} \]
where \( k \) is Coulomb's constant.

### Step 2: Analyze the Motion
As \( Q' \) moves in a circular path around \( Q \), the direction of the force \( F \) is always directed radially outward from \( Q \) to \( Q' \). The motion of \( Q' \) is tangential to the circular path.

### Step 3: Determine the Work Done
The work done \( dW \) on the charge \( Q' \) when it moves a small distance \( ds \) in the direction of the force \( F \) is given by:
\[ dW = F \cdot ds \]
However, since the force \( F \) is always directed outward and the displacement \( ds \) is tangential to the circular path, the angle \( \theta \) between the force and displacement is \( 90^\circ \).

### Step 4: Calculate the Work Done
Since the angle \( \theta \) is \( 90^\circ \), we have:
\[ \cos(90^\circ) = 0 \]
Thus, the work done for a small displacement becomes:
\[ dW = F \cdot ds \cdot \cos(90^\circ) = 0 \]
This means that the work done in moving \( Q' \) through any small segment of the circular path is zero.

### Step 5: Total Work Done for One Complete Revolution
Since the work done for each infinitesimal segment of the path is zero, the total work done \( W \) in making one complete revolution around the charge \( Q \) is:
\[ W = \int dW = 0 \]

### Conclusion
The work done on the charge \( Q' \) in making one complete revolution around the charge \( Q \) is zero.